Is x < y? from DS 1000 series

[mehravikas]

Question - Is x < y?

1. x - y + 1 < 0
2. x - y - 1 < 0

How would you go about solving this one?

[amitansu]

Ans A here.

From stem 1 : x-y+1<0>x-y<-1 i.e (x-y) negative i.e. x<y ;sufficient

From stem 2 : x-y-1<0>x-y<1 here it could be 0, more than zero or even negative.So not sufficient.

[mehravikas]

Ans A here.

From stem 1 : x-y+1<0>x-y<-1 i.e (x-y) negative i.e. x<y ;sufficient

From stem 2 : x-y-1<0>x-y<1 here it could be 0, more than zero or even negative.So not sufficient.

Hi,

I have a couple of more questions, which I have not answered correctly. Are you happy to answer more questions?

Vikas

[amitansu]

Go ahead and post 'em....


Amit

[mehravikas]

Go ahead and post 'em....


Amit

1. S is a set of integers such that

i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.

Is –4 in S?

(1) 1 is in S.
(2) 2 is in S.

2. If x and y are integers, is xy + 1 divisible by 3?

(1) When x is divided by 3, the remainder is 1.
(2) When y is divided by 9, the remainder is 8.

3. If x&#8800;0, is |x| < 1?
(1) x^2<1
(2) |x| < 1 / x

Thanks, for helping out Amit. Highly appreciate it.

Vikas

[amitansu]

Vikas,

Just a small request : Post one q per thread.So that it would be better to refer.

However for q i can you put it in a better way ? i couldn't follow it !!

For q 2 :

Ans would be E.

From 1 : x=3q+1 (q quotient here)

different values of x are 4,7,10,13 ......but no idea about y so insufficient

From 2: y=9q+8
so different values of y would be 17,26,35,44....still no idea about x . so insufficient.
Combining both also we have many combinations of xy which wouldn't verify whether (xy+1) is divisible by 3 oe not.Insufficient



For q 3 :

From 1 : x^2<1> +0r- (x)<1 which is mod(x)<1 here x being a factor.So sufficient.

From 2: mod(x)<1>+or-(x)<1/x (x is a factor here) but with -x being a factor would contradict the value.For example let x =1/2
so 1/2<1>1/2<2; but is -1/2 < 1/-(1/2) no.

However we are asked about whether mod(x)<1 ?
and from above we knew, yes definitely.So suficient.

Ans D.

[mehravikas]

Vikas,

Just a small request : Post one q per thread.So that it would be better to refer.

However for q i can you put it in a better way ? i couldn't follow it !!

For q 2 :

Ans would be E.

From 1 : x=3q+1 (q quotient here)

different values of x are 4,7,10,13 ......but no idea about y so insufficient

From 2: y=9q+8
so different values of y would be 17,26,35,44....still no idea about x . so insufficient.
Combining both also we have many combinations of xy which wouldn't verify whether (xy+1) is divisible by 3 oe not.Insufficient



For q 3 :

From 1 : x^2<1> +0r- (x)<1 which is mod(x)<1 here x being a factor.So sufficient.

From 2: mod(x)<1>+or-(x)<1/x (x is a factor here) but with -x being a factor would contradict the value.For example let x =1/2
so 1/2<1>1/2<2; but is -1/2 < 1/-(1/2) no.

However we are asked about whether mod(x)<1 ?
and from above we knew, yes definitely.So suficient.

Ans D.

Amit,

Thanks again. I'll post one question per thread. For question 2 the correct answer is given 'C'. Your answer to question 3 is correct. As for question 1, I am starting a new thread for it.

Vikas

[mehravikas]

1. S is a set of integers such that

i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.

Is –4 in S?

(1) 1 is in S.
(2) 2 is in S.

Hey Amit,

I have posted the q on a new thread, but I thought to give you some explanation on the q here. From above, the set should be something like -

S = {a, -a, b, ab}

1 is in S = {1, -1, b, 1 * b}
not sufficient to say whether -4 is in the set

2 is in S = {2, -2, b, 2b}
not sufficient to say whether -4 is in S

To my surprise the answer is 'B'. Let me know, if the explanation is of any help to you.

Vikas

[netigen]

(a) says that 1 is in the set. It does not say that only 1 is in the set. So the ans is -4 may or may not be in the set hence insufficient.

1. S is a set of integers such that

i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.

Is –4 in S?

(1) 1 is in S.
(2) 2 is in S.

Hey Amit,

I have posted the q on a new thread, but I thought to give you some explanation on the q here. From above, the set should be something like -

S = {a, -a, b, ab}

1 is in S = {1, -1, b, 1 * b}
not sufficient to say whether -4 is in the set

2 is in S = {2, -2, b, 2b}
not sufficient to say whether -4 is in S

To my surprise the answer is 'B'. Let me know, if the explanation is of any help to you.

Vikas

[mehravikas]

(a) says that 1 is in the set. It does not say that only 1 is in the set. So the ans is -4 may or may not be in the set hence insufficient.

1. S is a set of integers such that

i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.

Is –4 in S?

(1) 1 is in S.
(2) 2 is in S.

Hey Amit,

I have posted the q on a new thread, but I thought to give you some explanation on the q here. From above, the set should be something like -

S = {a, -a, b, ab}

1 is in S = {1, -1, b, 1 * b}
not sufficient to say whether -4 is in the set

2 is in S = {2, -2, b, 2b}
not sufficient to say whether -4 is in S

To my surprise the answer is 'B'. Let me know, if the explanation is of any help to you.

Vikas

What about point B {2 is in S}

[airan]

Vikas,
Reply to Question 2> Answer should be C and not E.
This is how:
X can be a number like 4, 7, 10, 13, 16, 19 etc ..
Y can be 17, 26 etc ..

1. is not sufficient.
2. this is also not sufficient because no relation to x.

Now combining both the statements,
if a number of form 3x+1 is multiplied by number of form 9x-1, it will always be divisible by 3 because the +1 will offset -1, try any of the above written combinations.
Hence answer is C.

[mehravikas]

Superb stuff dude !!

You have given me a tip to remember

[amitansu]

Yes, for Q 2 ans is "C" .
I just missed it !!

by pluggin the values of p and q we can confirm it !

Now for Q 1. i guess, since it says if some no. is there, then it's negation would also be there and as per 2 condition : if there are two nos. then their multiplication would also be there...

so, if 2 is there then -2 would also be there as per condition 1.
Again, if 2 and -2 are two existing nos. then their multiplication i.e. -4 should be there.So it meets the condition.

[airan]

That means u are considering both the conditions .. so answer cannot be B, it has to be C. B alone doesnt tells anything .!

[amitansu]

Considering the 2) of 1) & 2) not the first two statements (i) & (ii).